The volume of a parallelepiped with #vec(a)#, #vec(b)# and #vec(c)# as vectors coinciding with three edges sharing the same vertex equals to
#V_1 = vec(a) * (vec(b) xx vec(c))#
Let's examine what happens if we will use vectors #vec(b)+vec(c)#, #vec(c)+vec(a)# and #vec(a)+vec(b)# instead.
The volume of a new parallelepiped will be
#V_2 = (vec(b)+vec(c)) * ((vec(c)+vec(a)) xx (vec(a)+vec(b))) =#
# = (vec(b)+vec(c)) * (vec(c) xx vec(a)+vec(a) xx vec(a)+vec(c) xx vec(b)+vec(a) xx vec(b)) =#
# = vec(b) * (vec(c) xx vec(a))+vec(b) * (vec(a) xx vec(a)) + vec(b) * (vec(c) xx vec(b))+vec(b) * (vec(a) xx vec(b))+#
# + vec(c) * (vec(c) xx vec(a))+vec(c) * (vec(a) xx vec(a)) + vec(c) * (vec(c) xx vec(b))+vec(c) * (vec(a) xx vec(b)) #
As we know, vector product of collinear vectors equals to zero-vector. Therefore, #(vec(a) xx vec(a)) = vec(0)#.
Subsequent scalar product of any vector by zero-vector is zero.
Also, a scalar product of perpendicular vectors is equal to zero. Since #vec(b)# is perpendicular to #(vec(c) xx vec(b))#,
#vec(b) * (vec(c) xx vec(b)) = 0#
Analogously,
#vec(b) * (vec(a) xx vec(b)) = 0#
#vec(c) * (vec(c) xx vec(a)) = 0#
#vec(c) * (vec(c) xx vec(b)) = 0#
The result is:
#V_2 = vec(b) * (vec(c) xx vec(a))+vec(c) * (vec(a) xx vec(b))#
Each component of this sum equals to original volume of a parallelepiped since they are just cyclical permutation of vectors that preserves their order. Therefore, each is equal to #V_1#:
#V_2 = V_1 + V_1 = 40+40=80#
